Indian Forest Service (Mains), Chemistry optional · 2022 · Paper II
Q1(b). Cyanide addition to a hindered enone
Question
Predict the major and minor products of the enone MeCOC(Me)=C(Me)₂ with HCN, invoking kinetic and thermodynamic control. [structure of the enone given in the paper]
Answer in brief
Rapid 1,2-addition gives the cyanohydrin; reversible conditions favour the 1,4-addition product that retains the ketone.
Key points
- Carbonyl attack gives MeC(OH)(CN)C(Me)=C(Me)₂ after protonation: the kinetic 1,2-adduct.
- Attack at the β-carbon gives an enolate; protonation gives MeCOCH(Me)C(CN)(Me)₂, the conjugate-addition product.
- Cyanohydrin formation is reversible; under equilibrating conditions the carbonyl-containing 1,4-product is normally the thermodynamic major product.
- Low temperature and short reaction time can enrich the kinetic product.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coverskinetic vs thermodynamic control1,2 vs 1,4 additioncyanohydrinconjugate addition of HCNα,β-unsaturated ketone
All parts of Question 1
- 1(a) Pyridine versus imidazole5 marks
- 1(b) Cyanide addition to a hindered enone5 marks
- 1(c) Tracing the nitrile carbon5 marks
- 1(d) Dichlorocarbene trapping5 marks
- 1(e) Substituent effects on aryl negative charge5 marks
- 1(f) Why DMSO accelerates many SN2 reactions5 marks
- 1(g) Rearrangement during silver-assisted solvolysis5 marks
- 1(h) Organolithium and Baeyer–Villiger sequence5 marks